JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Gravitation

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A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height

h

is ______ s.

(A) $\frac{2 R _{e}}{g} [(1 + \frac{h}{R _{e}})^{3/2} - 1]$
(B) $\frac{1}{3} \frac{2 R _{e}}{g} [(1 + \frac{h}{R _{e}})^{3/2} - 1]$ ✓ Correct
(C) $\frac{1}{3} \frac{R _{e}}{2 g} [(1 + \frac{h}{R _{e}})^{3/2} - 1]$
(D) $\frac{R _{e}}{2 g} [(1 + \frac{h}{R _{e}})^{3/2} - 1]$

Answer: (B)

Solution

A body launched with exactly the escape speed has zero total energy, so at distance

r

from the centre of the earth

\frac{1}{2} v^{2} = \frac{GM}{r} \Rightarrow v = \frac{2 GM}{r} = R_{e} \frac{2 g}{r}

using

GM = g R_{e}^{2}

. Writing

v = \frac{d r}{d t}

and separating the variables,

\int_{R_{e}}^{R_{e} + h} r d r = R_{e} 2 g \int_{0}^{t} d t

\frac{2}{3} [(R_{e} + h)^{3/2} - R_{e}^{3/2}] = R_{e} 2 g t

t = \frac{2}{3} \frac{R _{e}}{2 g} [(1 + \frac{h}{R _{e}})^{3/2} - 1] = \frac{1}{3} \frac{2 R _{e}}{g} [(1 + \frac{h}{R _{e}})^{3/2} - 1]

The body takes

\frac{1}{3} \frac{2 R _{e}}{g} [(1 + \frac{h}{R _{e}})^{3/2} - 1]

to reach height

h

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Gravitation

Solution

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